I have the following question at hand and I have to admit that I am not used to integral remainder form of taylor approximation. I am still trying to work around, so a couple of hints would be useful before a full answer.

Basically I have to determine the constant $c$ such that:

$\int\limits_{0}^{x}f(t)dt=\int\limits_{x}^{1}t^2f(t)dt + \frac{x^8}{8}+\frac{x^6}{6}-\frac{c}{24}$

Thanks in advance!

EDIT: I am sorry, I forgot to mention that:

$f:[0,1]\to\mathbb R$

I suppose thats important. And f is continuous, although I would imagine that is implicit...

Now, I would like to know if there is any other trickier solution or any approach using the integral remainder form of taylor expansion. Thanks!

  • $\begingroup$ Have you been told what $f(t)$ is or are you suggesting that this equation holds for any function? Only it doesn't hold for arbitrary $f(t)$. $\endgroup$ – Graham Hesketh Aug 5 '13 at 18:07
  • $\begingroup$ The question seems somewhat incomplete. $\endgroup$ – André Nicolas Aug 5 '13 at 18:09
  • $\begingroup$ I am sorry, I forgot to mention that f is continous and from closed [0,1] to reals. See edit above. $\endgroup$ – user191919 Aug 5 '13 at 19:24
  • $\begingroup$ If you subs $x=0$ in the equation, you will get $\frac{c}{24}=-\int_{0}^{1} t^2f(t) dt$. $\endgroup$ – Mhenni Benghorbal Aug 5 '13 at 20:52
  • $\begingroup$ My attempt at the solution (and I suppose its right): With $x=0$, I got $\frac{c}{24}=\int_{0}^{1} t^2f(t) dt $ like Mhenni Benghorbal suggested. $\endgroup$ – user191919 Aug 6 '13 at 0:53

Plugging in $x=0$ and $x=1$ in the desired identity, one sees that $$ 24\int_0^1t^2f(t)\,\mathrm dt=c=7-24\int_0^1f(t)\,\mathrm dt. $$ There is no reason to expect that the LHS and the RHS coincide hence, in general, there is no $c$ such that the identity holds for every $x$ in $(0,1)$ (and in fact, for $x=0$ and $x=1$).

Finding $f$ such that the identity holds is another question: assuming that $f$ is a solution and differentiating with respect to $x$ yields $f(x)=-x^2f(x)+x^7+x^5$, hence $f(x)=x^5$. This is indeed a solution and, using the LHS of the identity above, one sees that $c=24\cdot\frac18=3$.

  • $\begingroup$ Neither Taylor expansion nor integral remainder here. $\endgroup$ – Did Aug 6 '13 at 13:00

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